Problem: Let N be a positive odd number.

There are N coins, numbered 1, 2, \ldots, N. For each i (1 \leq i \leq N), when Coin i is tossed, it comes up heads with probability p_i and tails with probability 1 - p_i.

Taro has tossed all the N coins. Find the probability of having more heads than tails.

Constraints

* N is an odd number.
* 1 \leq N \leq 2999
* p_i is a real number and has two decimal places.
* 0 < p_i < 1

Input

Input is given from Standard Input in the following format:


N
p_1 p_2 \ldots p_N


Output

Print the probability of having more heads than tails. The output is considered correct when the absolute error is not greater than 10^{-9}.

Examples

Input

3
0.30 0.60 0.80


Output

0.612


Input

1
0.50


Output

0.5


Input

5
0.42 0.01 0.42 0.99 0.42


Output

0.3821815872

Solution (in PYTHON):

Solution:
import decimal,math
# from decimal import Decimal as co
from collections import Counter,defaultdict
from fractions import gcd
from bisect import bisect_right,bisect_left
decimal.getcontext().prec = 10
import sys
sys.setrecursionlimit(1000000)
# co=decimal.Decimal
n=int(raw_input())
p = map(float,raw_input().split())
dp = [[(0.0)]*(n+1) for i in range(n+1)]
dp[0][0]=(1.0)
for i in range(1,n+1):
	for j in range(0,i+1):
		if(j==0):
			dp[i][j]=dp[i-1][j]*(1-(p[i-1]))
		else:
			dp[i][j]=dp[i-1][j]*(1-(p[i-1]))+dp[i-1][j-1]*(p[i-1])
ans=(0.0)
for j in range(n/2+1,n+1):
	ans=ans+dp[n][j]
print ans